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• In django models, how to make all table names not have the app label?

  - by Luigi

  I have a database that was already being used by other applications before i began writing a web interface with django for it. The table names follow simple naming standards, so the django model Customer should map to the table "customer" in the db. At the same time I'm adding new tables/models. Since I find it cumbersome to use app_customer every time i have to write a query (django's ORM is definitely not enough for them) in the other applications and I don't want to rename the existing tables, what is the best way to make all models in my django app use tables without applabel_, besides adding a Meta class with db_table= to each model? Is there any reason why I shouldn't do this? I have only one web app that needs to access this db, everything else doesn't use django models.

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• how to change display text in django admin foreignkey dropdown

  - by FurtiveFelon

  Hi all, I have a task list, with ability to assign users. So i have foreignkey to User model in the database. However, the default display is username in the dropdown menu, i would like to display full name (first last) instead of the username. If the foreignkey is pointing to one of my own classes, i can just change the str function in the model, but User is a django authentication model, so i can't easily change it directly right? Anyone have any idea how to accomplish this? Thanks a lot!

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• Incremement Page Hit Count in Django

  - by Andrew C

  I have a table with an IntegerField (hit_count), and when a page is visited (ie. http://site/page/3) I want record id 3 'hit_count' column in the database to increment by 1. The query should be like: update table set hit_count = hit_count + 1 where id=3 Can I do this with the standard Django Model conventions? Or should I just write the query by hand? I'm starting a new project, so I am trying to avoid hacks. We'll see how long this lasts! Thanks!

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• Django: cannot pass variable to included template?

  - by duy

  Hi, I got a problem where I want to use template including in Django. Here is the real example: I got 3 file: home.html (will get the context variable passed from Views), base.html (the skeleton template file) and the header.html (included by base.html). If if put the code below directly in base.html without including the header.html, the {{title}} variable passing from home is correctly called. But if I include the header.html in base.html, the {{title}} variable's value cannot be called. <title>{% block title %}{% endblock %} | {{ SITE_INFO_TITLE }}</title> Is there any solution to this problem? Thanks.

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• Django gives "I/O operation on closed file" error when reading from a saved ImageField

  - by Rob Osborne

  I have a model with two image fields, a source image and a thumbnail. When I update the new source image, save it and then try to read the source image to crop/scale it to a thumbnail I get an "I/O operation on closed file" error from PIL. If I update the source image, don't save the source image, and then try to read the source image to crop/scale, I get an "attempting to read from closed file" error from PIL. In both cases the source image is actually saved and available in later request/response loops. If I don't crop/scale in a single request/response loop but instead upload on one page and then crop/scale in another page this all works fine. This seems to be a cached buffer being reused some how, either by PIL or by the Django file storage. Any ideas on how to make an ImageField readable after saving?

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• I get a 400 Bad Request error while using django-piston

  - by Cheezo

  Hello, I am trying to use Piston to provide REST support to Django. I have implemented my handlers as per the documentation provided . The problem is that i can "read" and "delete" my resource but i cannot "create" or "update". Each time i hit the relevant api i get a 400 Bad request Error. I have extended the Resource class for csrf by using this commonly available code snippet: class CsrfExemptResource(Resource): """A Custom Resource that is csrf exempt""" def init(self, handler, authentication=None): super(CsrfExemptResource, self).init(handler, authentication) self.csrf_exempt = getattr(self.handler, 'csrf_exempt', True) My class (code snippet) looks like this: user_resource = CsrfExemptResource(User) class User(BaseHandler): allowed_methods = ('GET', 'POST', 'PUT', 'DELETE') @require_extended def create(self, request): email = request.GET['email'] password = request.GET['password'] phoneNumber = request.GET['phoneNumber'] firstName = request.GET['firstName'] lastName = request.GET['lastName'] self.createNewUser(self, email,password,phoneNumber,firstName,lastName) return rc.CREATED Please let me know how can i get the create method to work using the POST operation?

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• Correct redirect after posting comment - django comments framework

  - by Sachin

  I am using django comments framework for allowing users to comment on my site, but there is a problem with the url to which user is redirected after posting the comment. If I render my comment form as {% with comment.content_object.get_absolute_url as next %} {% render_comment_form for comment.content_object %} {% endwith %} Then the url to which it is redirected after the comment is posted is <comment.content_object.get_absolute_url/?c=<comment.id> For example I posted a comment on a post with url /post/256/a-new-post/ then the url to which I am redirected after posting the comment is /post/256/a-new-post/?c=99 where assume 99 is the id comment just posted. Instead what I want is something like this /post/256/a-new-post/#c99. Secondly when I do comment.get_absolute_url() I do not get the proper url instead I get a url like /comments/cr/58/14/#c99 and this link is broken. How can I get the correct url as mentioned in the documentation. Am i doing something wrong. Any help is appreciated.

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• Django adminsite customize search_fields query

  - by dArignac

  Howdy! In the django admin you can set the search_fields for the ModelAdmin to be able to search over the properties given there. My model class has a property that is not a real model property, means it is not within the database table. The property relates to another database table that is not tied to the current model through relations. But I want to be able to search over it, so I have to somehow customize the query the admin site creates to do the filtering when the search field was filled - is this possible and if, how? I can query the database table of my custom property and it then returns the ids of the model classes fitting the search. This then, as I said, has to flow into the admin site search query. Thanks!

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• Django User "per project" group assignation

  - by Ben G

  Hi, Here's my problem : my site has users, which can create projects, and access other user's projects. Each project can assign different rights to users. So, i could have Project A : user "John" is in group "manager" , and Project "B" user "John" is in group "worker". How could I use the Django User authentication model to do that ? From a SQL point a view, what i would like is to be able to add "project_id" in the primary key for the "auth_user_groups" table. I don't think profile is of any help here. Any advice ? UPDATE : "worker" and "manager" are just two examples of the permission group (or "roles") that my application defines. There will be more in the future. Eg : i will probably also have "admin", "reporting", etc...

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• Not work variables in django templates

  - by ??????? ???????

  My context dictionary not sending to my templates. I have function from django.shortcuts import render_to_response def home(request): return render_to_response('home.html',{'test':'test'}) and i have simple template such as: <html> <body> my test == {{test}} </body> </html> When i open my site in browser, i have "my test == ". settings.py is default. I dont use something custom. What the problem? Server is apache with wsgi module.

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• Django paging object has issues with Postgresql QuerySets

  - by pivotal

  I have some django code that runs fine on a SQLite database or on a MySQL database, but it runs into problems with Postgres, and it's making me crazy that no one has has this issue before. I think it may also be related to the way querysets are evaluated by the pager. In a view I have: def index(request, page=1): latest_posts = Post.objects.all().order_by('-pub_date') paginator = Paginator(latest_posts, 5) try: posts = paginator.page(page) except (EmptyPage, InvalidPage): posts = paginator.page(paginator.num_pages) return render_to_response('blog/index.html', {'posts' : posts}) And inside the template: {% for post in posts.object_list %} {# some rendering jazz #} {% endfor %} This works fine with SQLite, but Postgres gives me: Caught TypeError while rendering: 'NoneType' object is not callable To further complicate things, when I switch the Queryset call to: latest_posts = Post.objects.all() Everything works great. I've tried re-reading the documentation, but found nothing, although I admit I'm a bit clouded by frustration at this point. What am I missing? Thanks in advance.

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• Django - Passing arguments to models through ForeignKey attributes

  - by marshall

  I've got a class like this: class Image (models.Model): ... sizes = ((90,90), (300,250)) def resize_image(self): for size in sizes: ... and another class like this: class SomeClassWithAnImage (models.Model): ... an_image = models.ForeignKey(Image) what i'd like to do with that class is this: class SomeClassWithAnImage (models.Model): ... an_image = models.ForeignKey(Image, sizes=((90,90), (150, 120))) where i'm can specify the sizes that i want the Image class to use to resize itself as a argument rather than being hard coded on the class. I realise I could pass these in when calling resize_image if that was called directly but the idea is that the resize_image method is called automatically when the object is persisted to the db. if I try to pass arguments through the foreign key declaration like this i get an error straight away. is there an easy / better way to do this before I begin hacking down into django?

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• Problems using User model in django unit tests

  - by theycallmemorty

  I have the following django test case that is giving me errors: class MyTesting(unittest.TestCase): def setUp(self): self.u1 = User.objects.create(username='user1') self.up1 = UserProfile.objects.create(user=self.u1) def testA(self): ... def testB(self): ... When I run my tests, testA will pass sucessfully but before testB starts, I get the following error: IntegrityError: column username is not unique It's clear that it is trying to create self.u1 before each test case and finding that it already exists in the Database. How do I get it to properly clean up after each test case so that subsequent cases run correctly?

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• Django remove list_editable on a per row basis.

  - by Jason Leveille

  On the back of Django 1.2RC1, I have built a great administrator for my client! It's awesome. The client has one request that, if not satisfied, could bring this house of cards crashing down. I have a list of swim meet results. A meet result is added by a superuser. A team rep can edit a meet result (which they must be able to do inline, with list_editable), but after they edit the meet result inline and save, they should no longer be able to edit that result inline. They can only perform one edit. So, my question ... can I turn off list_editable on a per row basis?

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• Access to field in extended flatpage in django

  - by Stanislav Feldman

  How to access field in extended flatpage in django? I wrote this: class ExtendedFlatPage(FlatPage): teaser = CharField(max_length=150) class ExtendedFlatPageForm(FlatpageForm): teaser = CharField(max_length=150) class Meta: model = ExtendedFlatPage class ExtendedFlatPageAdmin(FlatPageAdmin): form = ExtendedFlatPageForm fieldsets = ( (None, {'fields': ('url', 'title', 'teaser', 'content', 'sites',)}), ) admin.site.unregister(FlatPage) admin.site.register(ExtendedFlatPage, ExtendedFlatPageAdmin) And creation in admin is ok. But then in flatpages/default.html I tried this: <html> <body> <h1>{{ flatpage.title }}</h1> <strong>{{ flatpage.teaser }}</strong> <p>{{ flatpage.content }}</p> </body> </html> And there was no flatpage.teaser! What is wrong?

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• (Django) Trim whitespaces from charField

  - by zardon

  How do I strip whitespaces (trim) from the end of a charField in Django? Here is my Model, as you can see I've tried putting in clean methods but these never get run. I've also tried doing name.strip(), models.charField().strip() but these do not work either. Is there a way to force the charField to trim automatically for me? Thanks. class Employee(models.Model): """(Workers, Staff, etc)""" name = models.CharField(blank=True, null=True, max_length=100) # This never gets run def clean_variable(self): data = self.cleaned_data['variable'].strip() return data def __unicode__(self): return self.name class Meta: verbose_name_plural = 'Employees' # This never gets run either class EmployeesForm(forms.ModelForm): class Meta: model = Employee def clean_description(self): #if not self.cleaned_data['description'].strip(): # raise forms.ValidationError('Your error message here') self.cleaned_data['name'].strip()

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• in django am facing url problem.....

  - by dpaksp

  am using django.0.97 version i have model called profile in that i have few fields eg like name ,email...ects and it's backend also ready..i.e database . and all users profile is created...i have given all permission to all users. when i login ,click on profile..i able to see list of all user name thr when i click on it ,it's goin to model page where i can edit the user profile..instead of that i want to navigate to a template when i can display the user details ,i have set the URl for it so that when url of that type request comes it should call a view from view it will call my template to display user details,.....the problem is it's not calling my view.... i think my problem is brief....if any information still required ?? pls ask me....and help me

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• Override Django inlineformset_factory has_changed() to always return True

  - by John

  Hi, I am using the django inlineformset_factory function. a = get_object_or_404(ModelA, pk=id) FormSet = inlineformset_factory(ModelA, ModelB) if request.method == 'POST': metaform = FormSet (instance=a, data=request.POST) if metaform.is_valid(): f = metaform.save(commit=False) for instance in f: instance.updated_by = request.user instance.save() else: metaform = FormSet(instance=a) return render_to_response('nodes/form.html', {'form':metaform}) What is happening is that if I change any of the data then everything works ok and all the data gets updated. However if I don't change any of the data then the data is not updated. i.e. only entries which are changed go through the for loop to be saved. I guess this makes sense as there is no point saving data if it has not changed. However I need to go through and save every object in the form regardless of whether it has any changes on not. So my question is how do I override this so that it goes through and saves every record whether it has any changes or not? Hope this makes sense Thanks

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• Django: query spanning multiple many-to-many relationships

  - by Brant

  I've got some models set up like this: class AppGroup(models.Model): users = models.ManyToManyField(User) class Notification(models.Model): groups_to_notify = models.ManyToManyField(AppGroup) The User objects come from django's authentication system. Now, I am trying to get all the notifications pertaining to the groups that the current user is a part of. I have tried.. notifications = Notification.objects.filter(groups_to_notify=AppGroup.objects.filter(users=request.user)) But that gives an error: more than one row returned by a subquery used as an expression Which I suppose is because the groups_to_notify is checking against several groups. How can I grab all the notifications meant for the user based on the groups he is a part of?

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• Django test client gets 301 redirection when accessing url

  - by Michal Klich

  I am writing unittests for django views. I have observed that one of my views returns redirection code 301, which is not expected. Here is my views.py mentioned earlier. def index(request): return render(request, 'index.html', {'form': QueryForm()}) def query(request): if request.is_ajax(): form = QueryForm(request.POST) return HttpResponse('valid') Below is urls.py. urlpatterns = patterns('', url(r'^$', 'core.views.index'), url(r'^query/$', 'core.views.query') ) And unittest that will fail. def so_test(self): response = self.client.post('/') self.assertEquals(response.status_code, 200) response = self.client.post('/query', {}) self.assertEquals(response.status_code, 200) My question is: why there is status 301 returned?

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• ModelName(django.contrib.auth.models.User) vs ModelName(models.Model)

  - by amr.negm

  I am developing a django project. I created some apps, some of those are related to User model, for instance, I have a feeds app that handles user feeds, and another app that deals with extra user data like age, contacts, and friends. for each of these, I created a table that should be connected to the User model, which I using for storing and authenticating users. I found two ways to deal with this issue. One, is through extending User model to be like this: ModelName(User): friends = models.ManyToMany('self') ..... Two, is through adding a foreign key to the new table like this: ModelName(models.Model): user = models.ForeignKey(User, unique=True) friends = friends = models.ManyToMany('self') ...... I can't decide which to use in which case. in other words, what are the core differences between both?

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• Django: where do I call settings.configure?

  - by RexE

  The Django docs say that I can call settings.configure instead of having a DJANGO_SETTINGS_MODULE. I would like my website's project to do this. In what file should I put the call to settings.configure so that my settings will get configured at the right time? Edit in response to Daniel Roseman's comment: The reason I want to do this is that settings.configure lets you pass in the settings variables as a kwargs dict, e.g. {'INSTALLED_APPS': ..., 'TEMPLATE_DIRS': ..., ...}. This would allow my app's users to specify their settings in a dict, then pass that dict to a function in my app that augments it with certain settings necessary to make my app work, e.g. adding entries to INSTALLED_APPS. What I envision looks like this. Let's call my app "rexe_app". In wsgi.py, my app's users would do: import rexe_app my_settings = {'INSTALLED_APPS': ('a','b'), ...} updated_settings = rexe_app.augment_settings(my_settings) # now updated_settings is {'INSTALLED_APPS': ('a','b','c'), 'SESSION_SAVE_EVERY_REQUEST': True, ...} settings.configure(**updated_settings)

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• Setting a preferred item of a many-to-one in Django

  - by Mike DeSimone

  I'm trying to create a Django model that handles the following: An Item can have several Names. One of the Names for an Item is its primary Name, i.e. the Name displayed given an Item. (The model names were changed to protect the innocent.) The models.py I've got looks like: class Item(models.Model): primaryName = models.OneToOneField("Name", verbose_name="Primary Name", related_name="_unused") def __unicode__(self): return self.primaryName.name class Name(models.Model): item = models.ForeignKey(Item) name = models.CharField(max_length=32, unique=True) def __unicode__(self): return self.name class Meta: ordering = [ 'name' ] The admin.py looks like: class NameInline(admin.TabularInline): model = Name class ItemAdmin(admin.ModelAdmin): inlines = [ NameInline ] admin.site.register(Item, ItemAdmin) It looks like the database schema is working fine, but I'm having trouble with the admin, so I'm not sure of anything at this point. My main questions are: How do I explain to the admin that primaryName needs to be one of the Names of the item being edited? Is there a way to automatically set primaryName to the first Name found, if primaryName is not set, since I'm using inline admin for the names?

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• Django admin proper urls inside listview

  - by hinnye

  Hi, My current target is to give users the chance to download CSV files from the admin site of my application. I successfully managed to create an additional column in the model's list view this way: def doc_link(self): return '<a href="files/%s">%s</a>' % (self.output, self.output) doc_link.allow_tags = True This shows the file name and creates the link, but sadly - because it's inside my 'searches' view - it has an URL: my_site/my_app/searches/files/13.csv. This is my problem, I would like to have my files stored in the admin media directory, like this: http://my_site/media/files/13.csv Does somebody know how to give url which points "outer" from the model's directory? Maybe somehow tell Django to use the ADMIN_MEDIA_PREFIX in the link? I'd really appreciate any help, thanks!

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• Saving data in a inherited django model

  - by aldeano

  I'm building an app to save data and some calculations made with those datas, the idea is keep the data in one model and the calculations in other. So, the models are like this: class FreshData(models.Model): name = models.CharField(max_length=20) one = models.IntegerField() two = models.IntegerField() def save(self, *args, **kwargs): Calculations() Calculations.three = self.one + self.two super(FreshData, self).save(*args, **kwargs) Calculations.save() class Calculations(FreshData): three = models.IntegerField() I've got a valueerror pointing out "self.one" and "self.two" as without value. I keep the idea in witch my design is wrong and django has a simpler way to store related data.

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